Compactness, connecteness and Hausdorffness on $ (S^2 /\mathscr{R}, \tau_{e_{|S^2}}/\mathscr{R}) $

Question

Consider over $S^2$ with the induced euclidean topology, the equivalence relation:

$(x,y,z)\mathscr{R}(x',y',z') \iff x+y=x'+y'$

Let $X=S^2 /\mathscr{R}$ be the quotient set with the quotient topology $\tau$, Prove that $(X,\tau)$ is compact, connected and Hausdorff. let $\tau_{e_{|S^2}}$ be the induced euclidean topology over the unit sphere $S^2$

My try:

For compactness I would say that since the sphere is closed and bounded, the property passes to the quotient , so the quotient $ (S^2 /\mathscr{R}, \tau_{e_{|S^2}}/\mathscr{R})=(X,\tau) $ is then compact

For connectedness since the sphere is path-connected, then the property passes to the quotient and it is connected.

I am unsure however if this reasonings are correct, because I haven't used the given $\mathscr{R}$, so this would imply it is valid for any equivalence relation

For Hausdorffness I am not sure how to proceed, since I think in this case the equivalence relation does comes into play.

$(S^2,\tau_{e_{|S^2})}$ is Hausdorff ,right?, because it is a subset of $\mathbb{R}^3$in a Hausdorff space

Provided we are on a Haussdorff space,I know that to prove the quotient is Hausdorff I have to prove that there exist disjoint saturated open neighbordhoods $A \in \mathscr{U}_p$, $B \in \mathscr{U}_q$ $\forall p , q$ not in the same equivalence class.

Can someone shed some light?

Answer 1

So first of all, I assume that $S^2=\{v\in\mathbb{R}^3\ |\ \lVert v\rVert=1\}$ is the standard 2-dimensional sphere. Well, it doesn't really matter what $S^2$ is, as long as it is a compact and connected subspace of $\mathbb{R}^3$. Recall that we have the standard projection

$$\pi:S^2\to S^2/\mathscr{R}$$ $$\pi(x)=[x]_\mathscr{R}$$

which is a continuous surjection.

For compactness I would say that since the sphere is closed and bounded, the property passes to the quotient , so the quotient $ (S^2 /\mathscr{R}, \tau_{e_{|S^2}}/\mathscr{R})=(X,\tau) $ is then compact

"Closed and bounded" property does not pass to quotients. Quotients need not be a subspace of $\mathbb{R}^n$, or even metrizable for "bounded" to make sense. Also quotient maps need not preserve closed sets. Nevertheless, the quotient will be compact.

We can use our projection $\pi$ and note that the image of a compact space is compact. Indeed, if $f:X\to Y$ is a continuous surjection with $X$ compact, then take $\mathscr{U}$ an open covering of $Y$. Then $f^{-1}(\mathscr{U})=\{f^{-1}(U)\ |\ U\in\mathscr{U}\}$ is an open covering of $X$. By compactness this covering has a finite subcovering $\{f^{-1}(U_1),\ldots,f^{-1}(U_n)\}$. You can easily check that $\{U_1,\ldots,U_n\}$ is a finite subcovering of $\mathscr{U}$.

For connectedness since the sphere is path-connected, then the property passes to the quotient and it is connected.

Yes. A continuous image of any (path) connected space is (path) connected, i.e. $\pi(S^2)=S^2/\mathscr{R}$ is (path) connected. The concrete formula for $\mathscr{R}$ is irrelevant.

For Hausdorffness I am not sure how to procede, since I think in this case the equivalence relation does comes into play.

It sure does. Unlike connectedness and compactness, not every quotient of a Hausdorff space is Hausdorff. But we have this neat property that a quotient of a compact Hausdorff space $X$ via relationship $R$ is Hausdorff if and only if $R$ is a closed subspace of $X\times X$. See this: Question about quotient of a compact Hausdorff space

So let's have a look at $\mathscr{R}=\{(x,y,z,x',y',z')\in\mathbb{R}^6\ |\ x+y=x'+y'\}$. Is that a closed subspace of $S^2\times S^2$? Sure it is, because we have a continuous function

$$f:S^2\times S^2\to\mathbb{R}$$ $$f(x,y,z,x',y',z')=x+y-x'-y'$$

and with that we have $\mathscr{R}=f^{-1}(\{0\})$ and so it is closed.